Breath Math

1. In the adjacent figure, two triangles are similar find the length of the missing side

Solution:

Let the triangles be ABC and PQR

^BC/_QR = ^AC/_PR

⁵/_X  = ¹³/₃₉

_{13X = 5 X 39}

_{X =} ^5X39/₁₃  = 5 X3 = 15

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2. What number is to 12 is 5 is 30?

Solution

Let x be the number

x:12 :: 5 : 30

30x = 12×5

x = ^12×5/₃₀   = 2

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3. Solve the following properties:

(i). x : 5 = 3 : 6

(ii) 4 : y = 16 : 20

(iii) 2 : 3 = y : 9

(iv) 13 : 2 = 6.5 : x

(v) 2 : π = x : ²²/₇

Solution:

(i). x : 5 = 3 : 6

6x = 5 x 3

6x = 15

x = ¹⁵/₆

 

(ii) 4 : y = 16 : 20

4×20 = 16y

y = ^4×20/₁₆

y = 5

 

(iii) 2 : 3 = y : 9

2×9 = 3y

y = ^2×9/₃ = 2×3 = 6

 

(iv) 13 : 2 = 6.5 : x

13x = 2 x 6.5

13x = 13

x = ¹³/₁₃ = 1

 

(v) 2 : π = x : ²²/₇

2x²²/₇ = πx

x = (2x²²/₇) /_π =(2x²²/₇) /(²²/₇)

x = 2

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4. find the mean proportion to :

(i) 8, 16

(ii) 0.3, 2.7

(ii)16²/₃ , 6

(iv)  1.25, 0.45

Solution:

(i) 8, 16

Let x be the mean proportion to 8 and 16

Then ⁸/_x = ^x/₁₆

x² = 8 x 16 = 128

x = √128 = √(64×2) = 8√2

 

(ii) 0.3, 2.7

Let x be the mean proportion to 0.3 and 2.7

Then ^0.3/_x = ^x/_2.7

x² = 0.3 x 2.7 = 0.81

x = √(0.81) = 0.9

 

(ii)16²/₃ , 6

Let x be the mean proportion to 16²/₃  and 6

Then (16²/₃) /_x = ^x/₆

x² = 16²/₃  x 6 = ⁵⁰/₃ x 6 = 100

x = √100 = 10

 

(iv)  1.25, 0.45

Let x be the mean proportion to 1.25 and 0.45

Then ^1.25/_x = ^x/_0.45

x² = 1.25 x 0.45

x = √(1.25 x 0.45) = √(1.25 x 0.45)x^{√(100×100)}/ _{√(100×100)}

= ^√(125×45)/_{√(100×100)} = ^{√(25x5x5x9)}/_{√(100×100)}  = ^5x5x3/_10×10  = ³/₄

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5. Find the fourth proportion for the following:

(i) 2.8,  14, 3.5

(ii) 3¹/₃, 1²/₃, 2¹/₂

(iii)1⁵/₇, 2³/₄, 3³/₅

Solution:

(i) 2.8,  14, 3.5

Let x be the  fourth proportion

Then, 2.8 : 14  :: 3.5  : x

2.8x = 14×3.5

x  = ^14×3.5/_2.8 = 17.5

 

(ii) 3¹/₃, 1²/₃, 2¹/₂

Let x be the  fourth proportion

Then, 3¹/₃: 1²/₃  :: 2¹/₂: x

¹⁰/₃ :⁵/₃ : : ⁵/₂ : x

¹⁰/₃x = ⁵/₃ x ⁵/₂

¹⁰/₃x = ²⁵/₆

x  = ²⁵/₆ x ³/₁₀ = ⁷⁵/₆₀  = ¹⁵/₁₂ = ⁵/₄

 

 

(iii)1⁵/₇, 2³/₁₄, 3³/₅

Let x be the  fourth proportion

Then, 1⁵/₇: 2³/₁₄:: 3³/₅: x

¹²/₇ :³¹/₁₄ : : ¹⁸/₅ : x

¹²/₇ x = ³¹/₁₄ x ¹⁸/₅

¹²/₇ x = ^31×18/_14×5

x  = ^31×18/_14×5 x ⁷/₁₂ = ^31×3/_5×4 = ⁹³/₂₀ = 4¹³/₂₀

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6. Find the third proportion to:

(i) 12, 16

(ii) 4.5, 6

(iii) 5¹/₂ , 16¹/₂

Solution:

(i) 12, 16

Let x be the third proportion

Then 16:12 :: x : 16

12x = 16 x16

x = ^16×16/₁₂ = ⁶⁴/₃  = 21¹/₃

 

(ii) 4.5, 6

Let x be the third proportion

Then 6:4.5 :: x : 6

4.5x = 6 x6

x = ^6×6/_4.5 = ³⁶/_4.5  = ³⁶⁰/₄₅  = 8

 

(iii) 5¹/₂ , 16¹/₂

Let x be the third proportion

Then 16¹/₂: 5¹/₂:: x : 16¹/₂

¹¹/₂ x = ³³/₂ x ³³/₂

x = ³³/₂ x ³³/₂ x ²/₁₁= ^33×3/₂  = ⁹⁹/₂  = 49¹/₂

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7. In a map ¹/₄ cm represents 25km, if two cities are 2¹/₂c apart on the map, what is the actual distance between them?

Solution:

Let 2¹/₂ cm represemts x km

¹/₄cm: 25km :: 2¹/₂cm : x km

¹/₄ x x = 25 x 2¹/₂

^x/₄ = ^{25 x5} /₂

x = ^25×5/₂ x 4 = 25x5x2 = 250km

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8. Suppose 30out of 500 components for a compter were found defective. At this aret how many defective components would he found in 1600 components?

Solution:

Number of defective components in 500 components = 30

Let x be the number of defecrive components in 1600 components

then 30:500 :: x :1600

30×1600 = 500x

x = ^30×1600/₅₀₀ =96

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